How do you solve #t^2 - t = 12 #?

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How do you solve #t^2 - t = 12 #?

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Answer 1 · 2016-07-31T19:53:55

#t=4#
#t=-3#

Explanation:

#t^2-t=12#
or
#t^2-t-12=0#
or
#t^2-4t+3t-12=0#
or
#t(t-4)+3(t-4)=0#
or
#(t-4)(t+3)=0#
or
#(t-4)=0#
or
#t=4#=======Ans #1#
or
#(t+3)=0#
or
#t=-3#=======Ans #2#

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How do you solve #t^2 - t = 12 #?

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